Cfrgtkky

I am trying to prove that if $R$ is a commutative Northerian ring with $1$, for all ideals $I$, $E$, then $E cap I^n subseteq EI$ for some $nin mathbb{N}$.

I have tried to go in this direction, but I have not obtained it:

Let $I= Q_{1}cap cdots cap Q_{m}$ be the primary decomposition of $I$ with $P_{i}=text{rad}(Q_{i})$, for all $iin {1,cdots,n}$ and $E=S_{1}cap cdots cap S_{t}$ a primary decomposition of $I$.

Since $R$ is Noetherian, for each $iin {1,cdots, m}, $ $Q_{i}^{n_{i}}subseteq P_{i}$ for some $n_{i}in mathbb{N}$.

Let $t=max{n_{i} mid iin {1,cdots,m}}$. Then, $Q_{i}^t subseteq P_{i}$ for all $iin {1,cdots,m}$. Therefore, $$(Q_{1}cap cdots cap Q_{m})^tsubseteq P_{i}subseteq Q_{i},$$ for $iin {1,cdots, m}$.

This is useless because I only obtain that $I^tsubseteq I$. However, I do believe that I should use the existence of primary decomposition in Noetherian rings. Can someone give me a clue, please?

Suppose that $EI= Q_{1}cap cdots cap Q_{n}$ where $Q_{i}$ is primary and $P_{i}=text{rad}(Q_{i})$ for $iin {1,cdots,n}$.

Suppose that $E$ is not contained in $Q_{i}$. Then, there exists $x$ in $E$ such that $x$ is not in $Q_{i}$. Let $zin I$. Then, $xzin EI=Q_{1}cap cdots cap Q_{n}$, so $xzin Q_{i}$ and therefore, $zin P_{i}$.

Since $R$ is a commutative Noetherian ring, there exists $k_{i}in mathbb{N}$ such that $P_{i}^{k_{i}}subseteq Q_{i}$.

Let $k=max{k_{i}mid iin{1,cdots,n}}$. Then,

$E subseteq bigcap_{Esubseteq Q_{i}} Q_{i}$ and $I^ksubseteq bigcap_{I^ksubseteq Q_{i}} Q_{i}$, so $Ecap I^ksubseteq EI$.